Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $z = \dfrac{-r - 10}{r^2 + r - 90} \times \dfrac{r - 9}{-5r + 35} $
Solution: First factor the quadratic. $z = \dfrac{-r - 10}{(r - 9)(r + 10)} \times \dfrac{r - 9}{-5r + 35} $ Then factor out any other terms. $z = \dfrac{-(r + 10)}{(r - 9)(r + 10)} \times \dfrac{r - 9}{-5(r - 7)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ -(r + 10) \times (r - 9) } { (r - 9)(r + 10) \times -5(r - 7) } $ $z = \dfrac{ -(r + 10)(r - 9)}{ -5(r - 9)(r + 10)(r - 7)} $ Notice that $(r + 10)$ and $(r - 9)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ -(r + 10)\cancel{(r - 9)}}{ -5\cancel{(r - 9)}(r + 10)(r - 7)} $ We are dividing by $r - 9$ , so $r - 9 \neq 0$ Therefore, $r \neq 9$ $z = \dfrac{ -\cancel{(r + 10)}\cancel{(r - 9)}}{ -5\cancel{(r - 9)}\cancel{(r + 10)}(r - 7)} $ We are dividing by $r + 10$ , so $r + 10 \neq 0$ Therefore, $r \neq -10$ $z = \dfrac{-1}{-5(r - 7)} $ $z = \dfrac{1}{5(r - 7)} ; \space r \neq 9 ; \space r \neq -10 $